If p, q, and r are different positive integers such that p + q + r = 6

MANHATTAN GMAT OFFICIAL SOLUTION:

Before beginning work on the statements, note that p, q, and r are different positive integers that add up to 6. Only one set of three numbers satisfies that condition: 1, 2, and 3. Therefore, in some order, p, q, and r are 1, 2, and 3.

(1) INSUFFICIENT: Three cases are possible:

The average of x and x^2 is x^3;
The average of x and x^3 is x^2;
The average of x^2 and x^3 is x.
The first case gives the equation {x+x^2}/2=x^3 , or x+x^2=2x^3. This is a factorable quadratic equation:

2x^3-x^2-x=0
x(2x^2-x-1)=0
x(2x+1)(x-1)=0

This equation gives three solutions for x: x = 0, x = –1/2, and x = 1. Because there are multiple possible values for x, the statement is not sufficient. (There’s no need to investigate the other possible cases at this point, since you’ve just determined that this statement is insufficient.)

(2) INSUFFICIENT: Just about any value for x will make this statement true (that the average of x^pxp and x^rxr is not x^qxq). For instance, if p = 1, q = 2, and r = 3, then the only requirement is that the average of x and x^3x3 not be equal to x^2x2. Try random values of x: If x = 2, then the average of x and x^3x3 is (2 + 8)/2 = 5, which is not equal to x^2=4x2 = 4. If x = 3, then the average of x and x^3x3 is (3 + 27)/2 = 15, which is not equal to x^2=9x2 = 9. Because there are multiple possible values for x, the statement is not sufficient.

(1) AND (2) INSUFFICIENT: In order to determine the values of x that satisfy both statements, revisit the cases derived in statement 1, this time solving for all of them.

Case #1: If the average of x and x^2 is x^3, then, as shown above, x can be any of 0, –1/2, 1.

Case #2: If the average of x and x^3 is x^2, then {x+x^3}/2=x^2, or x+x^3=2x^2. This is also a factorable equation: x^3-2x^2+x=0 ––> x(x^2-2x+1)=0 ––> x(x-1)^2=0. Therefore, x can be either 0 or 1.
Case #3: If the average of x^2 and x^3 is x, then {x^2+x^3}/2=x, or x^2+x^3=2x. This too is a factorable equation: x^3+x^2-2x=0 ––> x(x^2+x-2)=0 ––> x(x+2)(x-1)=0. Therefore, x can be any of 0, –2, 1.

There are thus four possible values of x according to statement 1: x = –2, –1/2, 0, 1.

For the values x = 0 or 1, the average of two terms is equal to the third term in all 3 cases, so these two values can’t satisfy statement 2 (where the average of the first two terms cannot equal the third term). Discard the values 0 and 1.

On the other hand, if x = –2, p and q are 2 and 3 (in either order), and r = 1, then both statements are satisfied. The average of x^p and x^q is x^r: (4 + -8)/2 = -2, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (4 + -2)/2 does not equal -8, so statement 2 is true. Therefore, x could equal –2.

What about the final case? If x = –1/2, p and q are 1 and 2 (in either order), and r = 3, then both statements are also satisfied. The average of x^p and x^q is x^r: (–1/2 + 1/4)/2 = –1/8, so statement 1 is true. In addition, the average of x^p and x^r is not x^q: (–1/2 + –1/8)/2 does not equal 1/4, so statement 2 is true. Therefore, x can still be either –2 or –1/2. Both statements together are still not sufficient to answer the question.

The correct answer is (E).